# Slant cone volume problem

I was given a a problem to solve, I thought I solved it but my answers don't look like the ones provided.

The Problem

A cone with radius of base r and height h, is stretched to the left and right by length a and b respectively such that height of the stretched cones equals to the height of the original cone and apexes of all the three cones are colinear.​

Devise an algorithm to calculate the volume of the union of the two stretched cones (colored blue in the image).

My first question was: what is the volume of a slant cone? What I've found so far is that the formula for the volume of a slant cone is the same as that of a regular cone (which made sense to me).

So I though the answer would be 2 times the area of a cone given by radius r and height h minus the intersection of the two slant cones, which is also a cone with radius r. I just need to find out the height of that smaller cone.

My thought was that I could just take the cross section of of the bodies and handle the problem in 2D. Where the base of the cone will lie on the x axis and the center for the circle lies in point (0,0).

What I did was calculate the equation of the line that goes from (-a,h) to (r,0). And I would also calculate the equation of the line going from (-r,0) to (b,h).

I proceeded to calculate the line equations and then to calculate where they intersect. I would then take y value at intersection and use it as the height of the smaller cone.

I really thought I nailed it since when I had a = 0 and b = 0 I got the volume of the original cone. However, when the answers where provided mine didn't match the ones given. Can someone point out the hole in my logic?

• I' not convinced (just at very first glance) that the intersection be a cone as well .. how do you reached at that? Dec 18, 2018 at 0:00
• @GCab I didn't prove it, which might me the problem in my attempt at reaching the solution. I just thought it natural, because if the ratio of the slopes is -1, I get an isosceles triangle, which, can then be turned into a cone by revolving around the y-axis. Dec 18, 2018 at 0:16
• so long as the height of the cone remains constant, volume does not change. The volume of an oblique cone is still the same as the non-oblique version. Think of the version with a right triangle. The area is always base times height, we don't care how stretched out it is so long as we know this data. Dec 18, 2018 at 0:26

For the principles of projection onto parallel planes, on a plane $$z=c$$, the section of the three cones are circles of same radius $$r(h-c)/h$$, and centers shifted in the same ratio $$c/h$$ wrt $$a,b$$.

So the intersection is a " conic wedge" structure: the superposition of the intersections of the shifted circles.

Then by Cavalieri principle, you can slant the cones symmetrically by $$(a+b)/2$$ wrt the right cone, compute the intersection of the two circles $$\cdot dz$$, and integrate.

• I think is a cone, but needs to be is like a cone, but —since the intersection of the circles isn't circular. (I spent a while trying to convince myself that the intersection somehow did manage to be conical.) Dec 18, 2018 at 1:20
• @timtfj Correct, the intersection is a series of increasing radius (downwards) intersecting circles where they completely overlap each other at the base. Dec 18, 2018 at 2:15

The volume of the intersection is given by the following integral based on a general equation for the area of the intersection of two same-sized circles in this scenario, dependent on the value of y:

$$V_i =\int_0^{h_i}\big(2\cos^{-1}(\frac{y(a+b)}{2r(h-y)})\cdot(r-\frac{ry}{h})^2 - \sqrt{(r-\frac{ry}{h})^2-(\frac{y(a+b)}{2h})^2}\cdot(\frac{y(a+b)}{h})\bigr) dy$$

Example: $$a=3, b=4, h=5\ \text{and}\ r=3$$

$$h_i = \frac{30}{13}$$ which is the height of the conic wedge

$$V_i =\int_0^{30/13} 2\cos^{-1}(\frac{7y}{6(5-y)})\cdot(3-\frac{3y}{5})^2 - \sqrt{(3-\frac{3y}{5})^2-(\frac{7y}{10})^2}\cdot(\frac{7y}{5}) dy$$

An integral solver couldn't come up with an anti-derivative but used an approximation method and came up with $$25.25762\ \#^3$$